SATHEE: Equilibrium Question 207

Equilibrium Question 207

Question: The solubility of $C a C O_{3}$ in water is $3.05 \times 10^{- 4} m o l e s / l i t r e .$ Its solubility product will be [MP PMT 1997]

Options:

A) $3.05 \times 10^{- 4}$

B) 10

C) $6.1 \times 10^{- 4}$

D) $9.3 \times 10^{- 8}$

Show Answer

Answer:

Correct Answer: D

Solution:

$\underset{S}{C a C O_{3}},$ ⇌ $\underset{S}{C a^{+ +}}, + \underset{S}{C O_{3}^{-, -}},$ Solubility product of $C a C O_{3}$

$K_{s p} = S^{2}$ ; $S = \sqrt{K_{s p}}$ It is a binary electrolyte. $S^{2} = K_{s p}$ ; ${(3.05 \times 10^{- 4})}^{2} = K_{s p}$ ; $K_{s p} = 9.3 \times 10^{- 8}$

Equilibrium Question 207

Question: The solubility of $C a C O_{3}$ in water is $3.05 \times 10^{- 4} m o l e s / l i t r e .$ Its solubility product will be [MP PMT 1997]

Options:

Answer:

Solution:

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Equilibrium Question 207

Question: The solubility of CaCO3 in water is 3.05×10−4moles/litre. Its solubility product will be [MP PMT 1997]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The solubility of $C a C O_{3}$ in water is $3.05 \times 10^{- 4} m o l e s / l i t r e .$ Its solubility product will be [MP PMT 1997]